Quotient Calculus

$ \def \hieruit {\quad \Longrightarrow \quad} \def \slechts {\quad \Longleftrightarrow \quad} \def \OF {\quad \mbox{or} \quad} $ So, if we eliminate $(2n-1)$ equations from the uniform three-diagonal system of linear equations, as defined in the preceding paragraph, then we have the following matrix patterns, but only for $n=2$ or $n=3$ : $$ \begin{array}{ccccccc} . & -a^2/(1-2ab) & 0 & 1 & 0 & -b^2/(1-2ab) & . \\ -a^3/(1-3ab) & 0 & 0 & 1 & 0 & 0 & -b^3/(1-3ab) \end{array} $$ So the quotient of the outer diagonal elements, with the variables $T_2$ and $T_4$ replaced by $T_1$ and $T_5$, for the case $ n=2 $, is $(b/a)^2$ .
And the quotient of the outer diagonal elements, with the variables $T_1,T_2,T_4,T_5$ eliminated, for the case $ n=3 $, is $(b/a)^3$ .
Thus for $ n=2 $ and $ n=3 $ we have that the quotient of the outer diagonals is $(b/a)^n$. After eliminating variables with the $ n=2,3 $ methods for the first time, all equations can be re-ordered to form two or three blocks in a new but equivalent tridiagonal system. This procedure can be repeated ad infinitum. It should be known from Multigrid Calculus that such a thing corresponds with grid coarsening, geometrically. Resulting in meshes which are coarsened two- or threefold every time again. The reverse procedure, corresponding with a two- or threefold mesh refinement, has taking the 2nd or 3rd root from the quotient as its algebraic equivalent. It is known from Multigrid Calculus that the $(b/a)^n$ law is even valid for powers $n=2^m$ where $m$ is a whole number, including zero. And for powers $n=2^{1/m}$ as well, where $m$ is a whole number except zero. Thus e.g the quotient of the outer diagonal elements for the case $ n=4 $ is still as could be conjectured from the cases $n=2$ and $n=3$, namely $b^4/a^4$. In very much the same way, it can be shown that the quotient law is valid for powers $n=3^m$ and for powers $n=3^{1/m}$. But, maybe somewhat unexpectedly, it can be proved with my favorite Computer Algebra System, MAPLE, that the conjecture breaks down at $n=5$:

  eliminate({-a*T0+T1-b*T2=0,-a*T1+T2-b*T3=0,-a*T2+T3-b*T4=0, 
             -a*T3+T4-b*T5=0,-a*T4+T5-b*T6=0,-a*T5+T6-b*T7=0, 
             -a*T6+T7-b*T8=0,-a*T7+T8-b*T9=0,-a*T8+T9-b*Ta=0}, 
             {T1,T2,T3,T4,T6,T7,T8,T9});

Giving, among a lot of garbage:

              3  3             2  2
         {20 b  a  T5 - 21 T5 a  b

              4     4    7     2                   6            5
         - 5 b  T5 a  + b  Ta a  + 8 b a T5 - 3 a b  Ta - T5 + b  Ta

              6         5       7     2
         - 3 a  b T0 + a  T0 + a  T0 b }]

Simplified by hand: \begin{eqnarray*} && (20 b^3 a^3 - 21 a^2 b^2 - 5 b^4 a^4 + 8 b a - 1) T_5 \\ && + (- 3 a^6 b + a^7 b^2 + a^5) T_0 + (- 3 a b^6 + a^2 b^7 + b^5) T_a = 0 \end{eqnarray*} Which proves that the quotient of the $(T_0,T_a)$ coefficients is not equal to $(b/a)^5$.
Needless to say that the quotient power law $(b/a)^n$ is also invalid for other numbers not being a power of $2$ and/or $3$, like for example $7$ or $11$.
Now imagine successive 2-fold or 3-fold grid coarsenings and refinements. We conclude that these multigrids correspond with an exponent $p$ in $(b/a)^p$ which must be equal to one of the following: $$ p = 3^n/2^m \OF p = 2^m/3^n $$ (Because it has no sense to multiply by 3 and then divide by 3 again.) Where $m$ and $n$ are whole numbers. It is not difficult to comprehend that any number $p$ can be approximated to arbitrary accuracy by a suitable choice of $m$ and $n$ .
Theorem. There exist natural numbers $m$ and $n$ such that for a given (positive) real number $p$ and a given uncertainity $\epsilon$ the following holds. $$ \left| p - \frac{3^n}{2^m} \right| < \epsilon \OF \left| p - \frac{2^m}{3^n} \right| < \epsilon $$ Proof. A well known theorem in number theory is that any irrational number $r$ can be approximated by a fraction $m/n$, where $(m,n)$ are naturals, such that: $$ \left| r - \frac{m}{n} \right| < \frac{1}{n^2} $$ This theorem can be proven with the Pigeonhole Principle. Or otherwise (with help of the Stern-Brocot tree) as exemplified in a 'sci.math' article titled Without the Pigeonhole Principle
The proof then reads as follows (apart from minor nitpicking). Consider the real number $r = \ln(x.3)/\ln(2)$: $$ \left| \frac{\ln(x.3)}{\ln(2)} - \frac{m}{n} \right| < \frac{1}{n^2} $$ $$ - \frac{1}{n^2} < \frac{\ln(x.3)}{\ln(2)} - \frac{m}{n} < \frac{1}{n^2} $$ $$ - \frac{\ln(2)}{n} < n \ln(x.3) - m \ln(2) < \frac{\ln(2)}{n} $$ $$ 2^{-1/n} < \frac{x^n.3^n}{2^m} < 2^{1/n} $$ Now determine $x$ such that $x^n=1/p$, which is done by $x=\exp(-\ln(p)/n)$: $$ 2^{-1/n} p < \frac{3^n}{2^m} < 2^{1/n} p $$ And determine $n$ from a given (relative) uncertainity $\delta$: $$ 2^{1/n} = (1+\delta) \slechts n = \frac{\ln(2)}{\ln(1+\delta)} $$ $$ 2^{-1/n} = \frac{1}{1+\delta} = \frac{(1-\delta)}{1-\delta^2} > 1-\delta $$ Therefore: $$ (1-\delta) p < 2^{-1/n} p < \frac{3^n}{2^m} < 2^{1/n} p = (1+\delta) p $$ $$ | p - \frac{3^n}{2^m} | < \delta p $$ At last, eventually replace the relative error by an absolute error, i.e. $\delta p = \epsilon$ , and we are done. By the way, then: $$ n = \frac{\ln(2)}{\ln(1+\epsilon/p)} \approx \frac{\ln(2)}{\epsilon/p} \hieruit n \approx \frac{p}{\epsilon} \qquad \mbox{(order of magnitude)} $$ Thus an estimate for the powers in $3^n/2^m$ is one divided by the allowed relative error in the approximation.
The above is the completion of a missing link in Multigrid Calculus. Up to now, I could only prove that $p$ should be of the form $2^n$ or 1$/2^n$ with no hope of extending this to arbitrary real exponents. The problem is stated on page 15 of the paper where it says "We seek to generalize" and then followed by an unsatisfactory solution.